Numerically computing $\int_0^1 \frac{1}{\sqrt{1-x^4}}dx$ In the book, "Pi and the AGM" by Borwein and Borwein, it is mentioned that Gauss computed the following integral to the eleventh decimal palce.
$\int_0^1 \frac{1}{\sqrt{1-x^4}}dx$
How did he do it?  Personnally, I looked at a Taylor expansion of
$\frac{1}{\sqrt{1-x}}$
Where I substituted $t^4$ for $x$, and integrated term by term, but this gives a series that converges really slowly.  Is there an obvious transform to make this computation faster?
 A: Here is one solution to compute $\int_0^1 \frac{1}{\sqrt{1-x^4}}dx$, with a big accuracy, without using arithmetic-geometric means.
The variable change x = sin u gives
$\int_0^1\frac{1}{\sqrt{1-x^4}}dx = \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1+sin^2 u}}du$
The second integral can be computed quickly using Simpson's rule, 8 intervals are enough for eleven decimal places.  Thanks to Joe Silverman for pointing me in the right direction.
A: A good place to look is pages 405 and 413 of the Nachlass section of Gauss's Werke III, which can be found online through Google Books.  On page 405, he gives the following formula for "$\text{arc sin lemn }x$":
$$\text{arc sin lemn }x= x+{1\over2}\cdot{1\over5}x^5 + {1\cdot3\over2\cdot4}{1\over9}x^9+{1\cdot3\cdot5\over2\cdot4\cdot6}{1\over13}x^{13}+{1\cdot3\cdot5\cdot7\over2\cdot4\cdot6\cdot8}{1\over17}x^{17}+\cdots$$
One page 413, he computes the value of $\int_0^1{dx\over\sqrt{1-x^4}}$, presumably using the expansion above, but explicitly citing the formula
$$\text{  arc sin lemn }{7\over23}+2\text{arc sin lemn }{1\over2}$$
obtaining
$$1.3110287771\quad460599052\quad320.7$$
He also notes there that Stirling had obtained the value $1.3110287771\ 4605987$.  This is in reference to the calculations on pages 57-58 of Stirling's Methodus differentialis from 1730, which can also be found through Google Books.  It might be worth noting that even Gauss was slightly off in the last couple of decimal places.  A more accurate value, which I took from here, is
$$1.3110287771\quad460599052\quad324197949$$
A: How Gauss (supposedly) calculated this integral in terms of AGM (namely $M(1,\sqrt{2})$) is outlined in http://home.sandiego.edu/~langton/gaussagm.pdf (Gauss, recurrence relations, and the AGM, by Stacy G. Langton). The final formula is $$\int\limits_0^1\frac{dx}{\sqrt{1-x^4}}=\frac{\pi}{2}\frac{1}{M(1,\sqrt{2})}.$$
A: I can't speak for Stirling, but if you break up the integral into an integral from 0 to 1/2 and an integral from 1/2 to 1, both of those converge very fast.
